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electrical_engineering_and_electronics:task_5u1zbroaz75w39jk_with_calculation [2024/07/15 15:18] – angelegt - Externe Bearbeitung 127.0.0.1electrical_engineering_and_electronics:task_5u1zbroaz75w39jk_with_calculation [2026/01/20 16:30] (aktuell) – mexleadmin
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 #@HiddenBegin_HTML~5u1zbroaz75w39jk_11,Path~@# #@HiddenBegin_HTML~5u1zbroaz75w39jk_11,Path~@#
  
-First, calculate the magnitude of the forces, like $\vec{F}_{01}$. \\+First, set up a coordinate system. Here, I choose x pointing to the right (positive values to the right) and y pointing upwards (positive values upwards). 
 + 
 +Then, calculate the magnitude of the forces, like $\vec{F}_{01}$ (force on q_0 from q_1). \\
 The force $\vec{F}_{01}$ is purely on the $x$-axis and therefore equal to $F_{01,x}$. The force $\vec{F}_{01}$ is purely on the $x$-axis and therefore equal to $F_{01,x}$.
 \begin{align*} \begin{align*}
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               = 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{VAs}\over{m}}}                = 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{VAs}\over{m}}} 
               = 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{Ws}\over{m}}} \\               = 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{Ws}\over{m}}} \\
-             &= 917.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the right)}+             &= +917.\;.\!.\!.\! {\rm \mu N}
 \end{align*} \end{align*}
 +Since both $q_0$ and $q_1$ have the same sign for their charges, they are repelling each other. \\
 +Therefore, The force $\vec{F}_{01}$ points to the right (positive value).
  
 Similarly, we get for $\vec{F}_{02}$ and $\vec{F}_{03}$ Similarly, we get for $\vec{F}_{02}$ and $\vec{F}_{03}$
 \begin{align*} \begin{align*}
-\vec{F}_{02} = F_{02,x} &= -1997.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the right)} \\ +\vec{F}_{02} = F_{02,x} &= -1997.\;.\!.\!.\! {\rm \mu N}  \\ 
-\vec{F}_{03} = F_{03,y} &= -1123.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the top)} \\+\vec{F}_{03} = F_{03,y} &= -1123.\;.\!.\!.\! {\rm \mu N} \\
 \end{align*} \end{align*}
 +Since $q_0$ and $q_2$ have the different sign for their charges, they are attract each other. 
 +Therefore, The force $\vec{F}_{02}$ points to the left (negative value). \\
 +Since $q_0$ and $q_3$ have the different sign for their charges, they are attract each other. 
 +Therefore, The force $\vec{F}_{03}$ points downwards (negative value). \\
  
  
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 #@HiddenBegin_HTML~5u1zbroaz75w39jk_12,Result~@# #@HiddenBegin_HTML~5u1zbroaz75w39jk_12,Result~@#
-  * $\vec{F}_{01} = \left( {\begin{array}{cccc}   917 {~\rm \mu N}  \\ 0                 \\ \end{array} } \right)$ +  * $\vec{F}_{01} = \left( {\begin{array}{cccc}   +917 {~\rm \mu N}  \\ 0                 \\ \end{array} } \right)$ 
-  * $\vec{F}_{02} = \left( {\begin{array}{cccc}  1997 {~\rm \mu N}  \\ 0                 \\ \end{array} } \right)$+  * $\vec{F}_{02} = \left( {\begin{array}{cccc}  -1997 {~\rm \mu N}  \\ 0                 \\ \end{array} } \right)$
   * $\vec{F}_{03} = \left( {\begin{array}{cccc}     0 {~\rm \mu N}  \\ -1123 {~\rm \mu N} \\ \end{array} } \right)$   * $\vec{F}_{03} = \left( {\begin{array}{cccc}     0 {~\rm \mu N}  \\ -1123 {~\rm \mu N} \\ \end{array} } \right)$
    
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 \begin{align*} \begin{align*}
 F = |\vec{F}| &= \sqrt{\left( \sum_i F_{i,x} \right)^2 + \left( \sum_i F_{i,y} \right)^2} \\ F = |\vec{F}| &= \sqrt{\left( \sum_i F_{i,x} \right)^2 + \left( \sum_i F_{i,y} \right)^2} \\
-              &= \sqrt{\left( +917 {~\rm \mu N} - 1997 {~\rm \mu N} \right)^2 + \left( 1123 {~\rm \mu N} \right)^2} \\+              &= \sqrt{\left( +917 {~\rm \mu N} - 1997 {~\rm \mu N} \right)^2 + \left( -1123 {~\rm \mu N} \right)^2} \\
               &= 1560.\;.\!.\!.\!  {~\rm \mu N} \\               &= 1560.\;.\!.\!.\!  {~\rm \mu N} \\
 \end{align*} \end{align*}